Answer to Question #113874 in Physical Chemistry for nobe

Question

Use oxidation numbers to identify the OA and RA, and to balance the following equation in acidic conditions.
NO_(3)^( -)(aq)+P_(4 )(s) ->H_(3)PO_(4)(aq)+NO_(2)(g)

Answer 1

Answer: Let's tabulate the oxidation numbers for each element in each molecule/ion.

NO₃ ^-(aq)+P₄(s) $\rightarrow$ H₃PO₄(aq)+NO₂(g)

NO₃^-: N(+5), O(-2)

P₄: P(0)

H₃PO₄: H(+1), P(+5), O(-2)

NO₂: N(+4), O(-2).

As you see, the oxidation number of P increased, and the oxidation number of N decreased. Therefore, the oxidizing agent is NO₃^- and the reducing agent is P₄.Let's write the half-reactions in acidic conditions:

P₄ + 16H₂O - 20e^- $\rightarrow$ 4H₃PO₄ + 20H⁺

NO₃^- + 2H⁺ + 1e^- $\rightarrow$ NO₂ + H₂O

Now we should add the two half-reactions, balancing the number of the electrons (to do this, multiply the second equation by 20):

P₄ + 16H₂O -20e^- + 20NO₃^- + 40 H⁺ + 20e^- $\rightarrow$ 4H₃PO₄ + 20 H⁺ + 20 NO₂ + 20H₂O

Simplifying:

P₄ + 20NO₃^- + 20 H⁺ $\rightarrow$ 4H₃PO₄ + 20 NO₂ +4H₂O.